Calculate the work done by a force vector F = y^2z^3i^+2xyz^3j^+3xy^2z^2k^ in moving an object along the path defined by the parametric equation x = t, y = t, z = t between the points A( t =0 ) and B( t = 1).
To calculate the work done by a force vector, we can use the formula:
W = ∫ F · dr
Where F is the force vector and dr is the infinitesimal displacement vector along the path. In this case, we need to calculate the dot product of the force vector F with the infinitesimal displacement vector dr and integrate it along the given path.
First, let's find dr, which represents the infinitesimal displacement vector. Since the path is defined by the parametric equations x = t, y = t, z = t, we can differentiate each equation with respect to t to find the components of dr:
dx = dt
dy = dt
dz = dt
Therefore, dr = dx i + dy j + dz k = dt i + dt j + dt k.
Now, let's calculate the dot product of F and dr:
F · dr = (y^2z^3)(dx) + (2xyz^3)(dy) + (3xy^2z^2)(dz)
= (y^2z^3)(dt) + (2xyz^3)(dt) + (3xy^2z^2)(dt)
= (y^2z^3 + 2xyz^3 + 3xy^2z^2) dt
Next, we will integrate the dot product over the given path from t = 0 to t = 1:
W = ∫(0→1) (y^2z^3 + 2xyz^3 + 3xy^2z^2) dt
To evaluate this integral, we need to substitute the given parametric equations into the expression inside the integral:
W = ∫(0→1) (t^2 * t^3 + 2t * t^3 + 3t * t^2 * t^2) dt
= ∫(0→1) (t^5 + 2t^4 + 3t^5) dt
= ∫(0→1) (4t^5 + 2t^4) dt
= [4/6 * t^6 + 2/5 * t^5] (0→1)
= (4/6 * 1^6 + 2/5 * 1^5) - (4/6 * 0^6 + 2/5 * 0^5)
= 4/6 + 2/5
= 20/30 + 12/30
= 32/30
= 16/15
Therefore, the work done by the force vector F in moving the object along the given path is 16/15.